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Old 2019-11-26, 05:43   #1
carpetpool
 
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"Sam"
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Post Gcd of consecutive factorial sums

Let s(n) = s(n-1) + n! where s(0) = 0. Then s(1) = 1, s(2) = 3, s(3) = 9, s(4) = 33, s(5) = 153, etc. This is the same as the sum of all factorials. Let

t(n) = gcd(s(n),s(n-1)) where n > 1. Then t(2) = 3, t(3) = 3, t(4) = 3, t(5) = 3, etc.

As n goes to infinity, does t(n) also go to infinity?

t(2) = 3
t(10) = 9
t(100) = 99
t(1000) = 99
t(2000) = 99

3 and 11 are the only primes below 2000 which t(n) can be divisible by.

What is the next prime such that t(n) can be divisible by (if it exists)?

Last fiddled with by carpetpool on 2019-11-26 at 05:44
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Old 2019-11-26, 11:47   #2
a1call
 
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Interesting question.
I don't have a complete proof but I think that no other primes than 3 and 11 can be common.
a!+b can only have primes less than a in common factors with a!.
So s(n) will always have only 3 and 11 in common prime factors with (n-2)!, which will propagate as n approaches to Infinity.

Last fiddled with by a1call on 2019-11-26 at 12:16
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Old 2019-11-26, 13:35   #3
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These sums were discussed a little over a year ago in this thread.

My contribution to this discussion, here.

One needs a prime p for which p divides 1! + 2! + ... + (p-1)!.

The primes p = 3 and p = 11 have this property.

I verified that there were no examples for 11 < p < 20000.

Update: Make that 11 < p < 100000.

I still don't have the foggiest notion of how to prove no further examples exist.
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Old 2019-11-26, 13:50   #4
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For prime p to become part of the gcd, it must divide the pth and p-1th term. Some interesting relations come out of it:

S(p-1) = S(p) - p! == 0 (mod p)
S(p-2) = S(p-1) - (p-1)! == 1 (mod p)
S(p-3) = S(p-2) - (p-2)! == 0 (mod p)
S(p-4) = S(p-3) - (p-3)! == 1/2 (mod p)
...
We can use (p-k)! == -1^k/(k-1)! (mod p) to keep going. Where that gets us, I don't know.

I see no reason why some p can't become part of the gcd. Naively, p should divide S(p-1) with probability 1/p, so it should just be a matter of turning over enough stones before one turns up.

BTW, checked up to p=500k. No solutions.
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Old 2019-11-26, 13:54   #5
R.D. Silverman
 
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Quote:
Originally Posted by Dr Sardonicus View Post
These sums were discussed a little over a year ago in this thread.

My contribution to this discussion, here.

One needs a prime p for which p divides 1! + 2! + ... + (p-1)!.

The primes p = 3 and p = 11 have this property.

I verified that there were no examples for 11 < p < 20000.

Update: Make that 11 < p < 100000.

I still don't have the foggiest notion of how to prove no further examples exist.
Under the assumption that the sum behaves as a random integer near (p-1)!,
the probability that a prime q < p divides the sum is 1/q. Now, sum over primes < n
of 1/q ~ log log n. This goes to oo. We should 'expect' log log(100000) ~ 2.44
primes less than 100000 to divide the sum.

As John Selfridge said: We know log log n goes to infinity. But no one has ever
observed it doing so.
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Old 2019-11-26, 15:56   #6
Dr Sardonicus
 
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Quote:
Originally Posted by R.D. Silverman View Post
Under the assumption that the sum behaves as a random integer near (p-1)!,
the probability that a prime q < p divides the sum is 1/q. Now, sum over primes < n
of 1/q ~ log log n. This goes to oo. We should 'expect' log log(100000) ~ 2.44
primes less than 100000 to divide the sum.

As John Selfridge said: We know log log n goes to infinity. But no one has ever
observed it doing so.
I couldn't think of any good reason more examples couldn't exist, so I find the heuristic appealing. Assuming it is anywhere close to being right...

Since exp(3)/log(10) = 8.72+ we're looking at maybe 9 decimal digits before finding a third example. So I'd say that finding a third example would require a much faster way to compute the required residue (mod p) than the one I was using.

Since exp(4)/log(10) = 23.71+ a fourth example might take quite a while.
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Old 2019-11-26, 17:28   #7
R.D. Silverman
 
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Quote:
Originally Posted by Dr Sardonicus View Post
I couldn't think of any good reason more examples couldn't exist, so I find the heuristic appealing. Assuming it is anywhere close to being right...

Since exp(3)/log(10) = 8.72+ we're looking at maybe 9 decimal digits before finding a third example. So I'd say that finding a third example would require a much faster way to compute the required residue (mod p) than the one I was using.

Since exp(4)/log(10) = 23.71+ a fourth example might take quite a while.
The same difficulty exists with Wieferich primes.
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Old 2019-11-26, 19:10   #8
carpetpool
 
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Quote:
Originally Posted by Dr Sardonicus View Post
The sums starting with 0! are even starting with k = 1, and greater than 2 for k > 1.

The sums starting with 1! are (as already observed) divisible by 3^2 for k > 4, and also by 11 for all k > 9.

The sum 0! + ... + 29! is 2*prime, and 1! + ... + 30! is 3^2 * 11 * prime.
s(500K) = 500K! + ... + 1! is divisible by 99 and t(500K) = 99.

s(n)/99 is prime for (n>10) n=:

11, 13, 29, 31, 47, 54, 79, 87, 103, 107, 177, 191, ... (up to 200)

It's unclear if s(n)/99 is prime infinitely often as this depends on the existence of another prime p dividing s(p) and s(p+1). It seems likely to exists by the heuristics R.D.S. mentioned.

2 is the only prime that divides s(n)+1 (left factorial sum) infinitely up to n = 2000.

92504 = 2^3*31*373 divides s(n)-1 infinitely up to n = 2000.

The primes dividing any generalized factorial sum in particular seem to be random.
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Old 2019-11-26, 19:28   #9
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Quote:
Originally Posted by carpetpool View Post
s(500K) = 500K! + ... + 1! is divisible by 99 and t(500K) = 99.

s(n)/99 is prime for (n>10) n=:

11, 13, 29, 31, 47, 54, 79, 87, 103, 107, 177, 191, ... (up to 200)

It's unclear if s(n)/99 is prime infinitely often as this depends on the existence of another prime p dividing s(p) and s(p+1). It seems likely to exists by the heuristics R.D.S. mentioned.

2 is the only prime that divides s(n)+1 (left factorial sum) infinitely up to n = 2000.

92504 = 2^3*31*373 divides s(n)-1 infinitely up to n = 2000.

The primes dividing any generalized factorial sum in particular seem to be random.
Off topic for this sub-forum. Will a moderator please move it?
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Old 2019-11-26, 22:03   #10
carpetpool
 
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Quote:
Originally Posted by R.D. Silverman View Post
Off topic for this sub-forum. Will a moderator please move it?
I meant to put that on my own thread...
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Old 2019-11-26, 22:20   #11
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Quote:
Originally Posted by carpetpool View Post
I meant to put that on my own thread...
Which thread. A mod can move the post.
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